“微积分二：数列与级数”将介绍数列、无穷级数、收敛判别法和泰勒级数。本课程不仅仅满足于得到答案，而且要做到知其然，并知其所以然。

Loading...

From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

45 ratings

“微积分二：数列与级数”将介绍数列、无穷级数、收敛判别法和泰勒级数。本课程不仅仅满足于得到答案，而且要做到知其然，并知其所以然。

From the lesson

交错级数

在第四个模块中，我们讲解绝对和条件收敛、交错级数和交错级数审敛法，以及极限比较审敛法。简而言之，此模块分析含有一些负项和一些正项的级数的收敛性。截至目前为止，我们已经分析了含有非负项的级数；如果项非负，确定敛散性会更为简单，因此在本模块中，分析同时含有负项和正项的级数，肯定会带来一些新的难题。从某种意义上，此模块是“它是否收敛？”的终结。在最后两个模块中，我们将讲解幂级数和泰勒级数。这最后两个课题将让我们离开仅仅是敛散性的问题，因此如果你渴望新知识，请继续学习！

- Jim Fowler, PhDProfessor

Mathematics

Irrationality.

[MUSIC]

What is an irrational number?

Well, here's a definition.

A real number x is irrational if it can't be expressed as a rational number.

If it can't be written as p over q for p and q integers.

You may have already met some irrational numbers.

For example, here's a claim, the square root of 2 is irrational.

How do I go about proving such a claim?

Well, the argument begins by supposing

that I can write the square root of 2 as a rational number.

And if I can write the square root of 2 as a rational number,

then I can write the square root of 2 as a fraction in lowest terms.

So I'm supposing that a and b don't have any factors in common.

All right, now once I've written the square root of 2 as a/b,

I could square both sides and then I'd find that 2 is a squared/b squared.

And then I could multiply both sides of this equation by b squared, and

I'd find that 2b squared = a squared.

Now what does that mean?

Well, that means, in particular, that a is a multiple of 2, right?

So that means that a must be even, so I can write a as 2 times some integer.

I'll call that integer K.

So a is even, a is twice some integer.

Now I can plug this fact back into here, right?

Instead of saying 2b squared is a squared,

I can say that 2b squared is now a, which is 2 times an integer squared.

And if I expand that out, I get that that's 2b squared is 4K squared.

Now if I divide both sides of this by 2,

then I get that b squared is 2K squared.

And what does that mean about b?

Well that means that b is also a multiple of two.

That means that b is even.

But if a and b are both even,

then the fraction a over b is not in lowest terms, right?

I've got a common factor of two in the numerator and denominator.

And that's the contradiction

that then shows that the square root of two can't be written as a fraction.

That's a pretty standard method for showing that a number's irrational.

You start out by assuming that it's a rational number, and

then derive some kind of contradiction which reveals that your original

assumption that the number was rational must have been an error.

And then you know the number's irrational.

I'll try the same kind of game, but not with the square root of 2 but

with the number e.

Well, if e were a rational number than its reciprocal, 1/e,

would be a rational number as well.

If I could write e as a fraction,

this would just be the reciprocal of that fraction.

I can analyze 1/e, turns out that 1/e

is equal to the value of this convergent and alternating series.

Now the quality of 1/e with the value of this series

isn't something that we're quite able to do.

We're going to see that these are equal in week six of this course.

So there's a little bit more to do here, all right.

But here's the big deal.

I can analyze this alternating series.

So our goal is going to be to show that the value of this alternating series is

irrational, and at some future point, we're going to see that the value of this

series is 1/e, which will then mean that e also must be irrational.

For the meantime,

let's just show that the value of that series is an irrational number.

So let's suppose that the value of this series,

the sum n goes from 0 to infinity of -1 to the n over n factorial.

Let's suppose that this is equal to a rational number,

that it is a/b for integers a and b.

Now let's look at the bth partial sum.

Yeah, I really mean this b.

The b from the denominator of this fraction that I'm imagining 1/e is

equal to.

So let's take a look at that bth partial sum, S sub b, right.

It's just the sum of the terms.

n goes from zero to b of -1 to the n, over n factorial.

Now, the whole deal here is that I've got an alternating series.

And what that means to the true value of the alternating series, which is 1/e,

must be between the S sub b and S sub b+1 partial sums, right?

That's the trick about the error bounds for an alternating series, that the true

value of the alternating series is always between neighboring partial sums.

Now the significance of this is that I know exactly how far apart S sub b and

S sub b+1 are.

S sub b and S sub b+1 are exactly 1/b+1 factorial apart,

because S sub b+1 and S sub b differ by the b+1th term in the series.

So I could put these two statements together to make this claim.

That the distance between 1/e,

the true value of the series and the s sub b partial sum,

the distance between these two numbers is no more than this, 1 over b + 1 factorial.

Now, I multiplied both sides by b factorial, right.

So I've got this inequality.

Multiplying both sides of this inequality by the positive number b factorial.

And what do I have?

Well, I've got b factorial over b+1 factorial.

This side can be simplified a bit.

This is 1 over b+1.

And one thing I know about 1 over b + 1 is that it's less than 1.

I can also simplify the left hand side.

Well first of all, I know that 1/e isn't actually equal to the partial sum.

This is the true value of the series.

This is just the sum of the first b terms.

These are not the same.

So, that tells me that I've got 0 less than this quantity.

Now I can also simplify this a little bit more.

I can say that this is b factorial times, what's 1/e?

Well assuming that I can write 1/e as this fraction,

a over b, and I could distribute this to get this, right?

b factorial times 1/e, which is, I'm assuming, a/b,

minus b factorial times the bth partial sum.

But I also know about something about the quantity that I'm taking

the absolute value of.

Specifically, I know that b factorial times a/b is an integer, right?

This is just b- 1 factorial times a.

That's an integer.

I also know something about b factorial times the bth partial sum.

All right, what is that?

Well, by definition, that's just b factorial, and

here I've written out the bth partial sum.

But, now I could put the b factorial inside there, and I'd get that this is

the sum n goes from zero to b of -1 to the n + 1 times b factorial over n factorial.

And what do I know?

b is at least as big as n, so b factorial divided by n factorial, that's an integer.

This is the sum and difference of integers.

That means that b factorial times the bth partial sum is an integer,

and that is problematic.

Because if this is an integer, and this is an integer,

that means this quantity inside here is also an integer.

That means that there's some integer which is between 0 and 1.

But that is not true.

Well, since there is no integer between 0 and 1, that's a contradiction,

and consequently, it must be the case that 1/e is irrational.

And therefore, e is irrational, which is what I wanted to show.

I still owe you a proof that that alternating series evaluates to 1/e,

and that proof is coming.

But in the meantime, it's worth reflecting on why this argument worked at all.

The cool thing about alternating series is that they come with error bounds,

I know that the true value of an alternating series

is between neighboring partial sums.

And that's great for doing numerics.

That's great for doing computation.

But the real point of computation isn't numbers, it's insight.

And I think this argument, this proof that 1/e and therefore e is irrational,

is a really great example of how a computational tool,

the fact that we have these explicit error bounds, can yield real insight.

[NOISE]

Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.